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Hexagonally closer (Posted on 2019-07-07) Difficulty: 4 of 5
There is a regular hexagon whose inradius in 1 unit. A region is created inside the hexagon which is made up of all the points that are closer to the center of the hexagon than any of its sides. Find the area of this region.

No Solution Yet Submitted by Danish Ahmed Khan    
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Some Thoughts A start | Comment 1 of 3
For purposes of determining the boundary of the sought area, it would be best to consider only one equilateral triangle that makes up 1/6 of the whole hexagon. That avoids discontinuities that would distract (and detract) from identifying the type of boundary it is.

The center of the hexagon is the focus of a parabola and the opposite side of the chosen equilateral triangle is a segment of a line that serves as the directrix of the parabola that forms the boundary of (the portion of) the area asked for.

The fraction of the triangle within the parabola between the bounding sides of the triangle is repeated for the other five triangles that make up the hexagon. Now, the goal is to find the area of the slice-of-pie shaped sector of the parabola.

  Posted by Charlie on 2019-07-07 09:57:04
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